Answer: \(\theta=15.2^\circ,\ 44.8^\circ,\ 135.2^\circ,\ 164.8^\circ\).

We start with the main method. Rewrite everything in terms of \(\sin x\) and \(\cos x\), then simplify using standard identities.

For part (a), write \(\tan x\) and \(\sec x\) in terms of sine and cosine:

\(\tan x+\sec x=\frac{\sin x}{\cos x}+\frac{1}{\cos x} =\frac{1+\sin x}{\cos x}.\)

Therefore

\((\tan x+\sec x)^2=\frac{(1+\sin x)^2}{\cos^2x}.\)

Since \(\cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x)\),

\((\tan x+\sec x)^2=\frac{(1+\sin x)^2}{(1-\sin x)(1+\sin x)}.\)

Hence

\((\tan x+\sec x)^2=\frac{1+\sin x}{1-\sin x}.\)

For part (b), apply this identity with \(x=3\theta\):

\(\frac{1+\sin3\theta}{1-\sin3\theta}=6.\)

Let \(u=3\theta\). Then

\(1+\sin u=6(1-\sin u).\)

So

\(1+\sin u=6-6\sin u\), giving

\(7\sin u=5.\)

Thus

\(\sin u=\frac57.\)

Since \(0^\circ\leqslant\theta\leqslant180^\circ\),

\(0^\circ\leqslant u\leqslant540^\circ.\)

The values of \(u\) in this interval are approximately

\(45.58^\circ,\ 134.42^\circ,\ 405.58^\circ,\ 494.42^\circ.\)

Divide by \(3\):

\(\theta=15.2^\circ,\ 44.8^\circ,\ 135.2^\circ,\ 164.8^\circ.\)

This completes the solution and gives the required result.